document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1148882>Question 1148882</A>: <I><SPAN STYLE=\"word-break: break-all;\"> The denominator of a fraction is five more than twice the numerator. If both numerator and denominator are decreased by seven, the simplified result is 1/6.<BR>\n" );
document.write( ". Find the original fraction. (Do NOT simplify.) </SPAN>\n" );
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document.write( "Answer: <font size=4 color=red>10/25</font><BR>\n" );
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document.write( "Work Shown:\r<BR>\n" );
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document.write( "Let<BR>\n" );
document.write( "n = original numerator<BR>\n" );
document.write( "d = original denominator\r<BR>\n" );
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document.write( "We know that d = 2n+5 since \"The denominator of a fraction is five more than twice the numerator\"\r<BR>\n" );
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document.write( "\"If both numerator and denominator are decreased by seven, the simplified result is 1/6\" meaning that (n-7)/(d-7) = 1/6\r<BR>\n" );
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document.write( "Let's solve for n after doing a substitution<BR>\n" );
document.write( "(n-7)/(d-7) = 1/6<BR>\n" );
document.write( "(n-7)/(2n+5-7) = 1/6 ... replace d with 2n+5<BR>\n" );
document.write( "(n-7)/(2n-2) = 1/6<BR>\n" );
document.write( "6(n-7) = 1(2n-2) .... cross multiply<BR>\n" );
document.write( "6n-42 = 2n-2<BR>\n" );
document.write( "6n-42-2n = 2n-2-2n ... subtract 2n from both sides<BR>\n" );
document.write( "4n-42 = -2<BR>\n" );
document.write( "4n-42+42 = -2+42 ... add 42 to both sides<BR>\n" );
document.write( "4n = 40<BR>\n" );
document.write( "4n/4 = 40/4 ... divide both sides by 4<BR>\n" );
document.write( "n = 40/4<BR>\n" );
document.write( "n = 10\r<BR>\n" );
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document.write( "If n = 10, then d is<BR>\n" );
document.write( "d = 2n+5<BR>\n" );
document.write( "d = 2*10+5<BR>\n" );
document.write( "d = 20+5<BR>\n" );
document.write( "d = 25\r<BR>\n" );
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document.write( "So n/d = <font color=red>10/25</font> is the original fraction (we arent reducing it). After decreasing the numerator and denominator by 7, we end up with 3/18 = 1/6, which confirms our answer.<BR>\n" );
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